A Problem Booklet for IIM-CAT 2008

School of Management Quantitative Aptitude (Volume IV)

A Sample Booklet( not for sale)

Rahul Kumar Jha

Mar-Nov 2008 Foreword

This is the fourth set in the series. hope you had fun in the first three sets,like the third set this set also has solutions and keys, kindly check at the end.I have taken proper care to select nice problems and tried to give one of thebest if not the best solution. Still if you can’t understand teh solution or youhave a better approach, you know how to find me.

Problem Set 4

201) How many zeros at the end of binary expansion of 100!?

a) 20 b) 24 c) 50 d) 75 e) 97

202) How many number are there in range of 10 to 106 such that their digitsum is 18 ? For ex. 99000, 9072, 90027, 66006.. .

212) Find the number of pairs of positive integers (a, b) such that a! + b! = ab

2a) 0 b) 1 c) 2 d) 3 e) none of these

213) Ganesh and sarath walk up an escalator which moves up with a con-stant speed. For every two steps that ganesh takes, sarath takes one step.Ganesh gets to the top of the escalator after having taken 30 steps, whilesarath takes only 20 steps to reach the top.If the escalator were turned off,how many steps should each o them take to reach the top?

a) 32008 b)32009 c) 32010 d) 32011 e) Can not be uniquely

216) Find the no of triplets (x, y, z) of positive integers such that xy + yz +

zx = xyz

a) 0 b) 1 c) 4 d) 7 e) 10

3217) In the figure shown above, ABCD is a square alongwith two othersquares which are inscribed in ABCD. Find the ratio of the area of S to thearea of ABCD √a) 25/54 b) 71/81 c) 30/81 d) 10/9 e) none of these

218) If the lcm of (10!)(18!) and (12!)(17!) is expressed as a!b!/c! where a, b

are two digit positive numbers and c is a one digit positive number. Thenabc equals ?

a) 54 b) 648 c) 216 d) 324 e) none of these

219)Find the remainder when 2184 is divided by 343

a) 0 b) 1 c) 49 d) 341 e) none of these

Directions for problem 220-221:Each question is followed by 2 state-

ments X and Y. Answer each question using the following instructionsChoose A if the question can be answered using X aloneChoose B if the question can be answered using Y aloneChoose C if the question can be answered using either X or (exclusive) YChoose D if the question can be answered using X and Y togetherChoose E if the question can not be answered using X and Y also

222)Two friends get home at 2:30 P.M. They both go to bed 10 P.M. Ifboth friends will go to a park twice for 15 minutes, what is the probabilitythat they will be able to see each other at any random time between the twotime intervals? Assume that the travel to and from the park takes no time. 25 8 27 29 3(A) (B) (C) (D) (E) 225 15 225 225 17223) Let p be a prime number such that f (p) = (2 + 3) − (22 + 32 ) + (23 +33 ) − ... − (2p−1 + 3p−1 ) + (2p + 3p ) is divisible by 5. If p1 , p2 and p3 are thethree smallest such primes, then p1 + p2 + p3 equals

230) Amit has a total of n balls of which r, g, band w are red, green, blueand white respectively. His friend Sumit asks him about the balls he says,I have so many balls such that if I select four balls, probability of selectingfour red balls is same as the probability of one ball of each kind and is sameas probability of selecting one white and three red balls and is same as theprobability of selecting two red, one white and one blue ball. Sumit asks himwhether this is the minimum of such set of balls. Amit says yes. He tellshim the value of r. What is it?

a) 9 b) 10 c) 11 d) 12 e) 13

231) Amit asks the total number of balls, sumit answers correctly what is hisanswer?

a) 17 b) 19 c) 21 d) 23 e) 25

232) Given that x2 + y 2 = 14x + 6y + 6, what is the largest possible value

that 3x + 4y can have?

(A) 72 (B) 73 (C) 74 (D) 75 (E) 76

6233) The director of a marching band wishes to place the members intoa formation that includes all of them and has no unfilled positions. If theyare arranged in a square formation, there are 5 members left over. The di-rector realizes that if he arranges the group in a formation with 7 more rowsthan columns, there are no members left over. Find the maximum numberof members this band can have.

235) The perimeter of a right triangle is 60. The height to the hypotenuse is12 what is the area?

(A) 75 (B) 144 (C) 150 (D) 300 (E) noneof these

236) MN is the diameter of a semicircle, the full circle has center O. A

and B are on the semicircle and C is a point on ON such that minor arc MA= 40 degrees, angle OAC = 10 degrees, and angle OBC is 10 degrees. findthe degree measure of minor arc BN.

(A) 10 (B) 20 (C) 30 (D) 40 (E) none of these

237) Given that A be a set such that A={1,2,A}. Then n(A) equals?

(A) 2 (B) 3 (C) 4 (D) can not be determined (E) none of these

238) 4 identical dice are tossed simultaneously. What is the probability thatat least 3 of the four nos. shown are different? 20 120 5 1200(A) (B) (C) (D) (E) none of these 216 216 6 1296239) In triangle ABC the medians AM and CN to sides BC and AB, respec-tively, intersect in point O. P is the midpoint of side AC, and M P intersects

7CN in Q. If the area of triangle OM Q is n, then the area of triangle ABC is:

(A) 16n (B) 18n (C) 21n (D) 24n (E) 27n

240) x, y, z, and w are real numbers. If x, y, z lies in interval of (1, ∞) and

w is in interval of (0, ∞), and that:

logx w = 24logy w = 40logxyz w = 12Find the value of logz w.

(A) 16 (B) 24 (C) 36 (D) 48 (E) 60

241) Let A and B be opposite vertices of a cube. A round-trip path is

defined to be a path on the cube from A to B and then back to A, using theleast possible number of edges, and without using the same edge more thanonce. How many round-trip paths are there?

(A) 6 (B) 12 (C) 18 (D) 24 (E) 36

242) Find the number of pairs of positive integers (x, y) such that

xy = y 50

(A) 6 (B) 7 (C) 8 (D) 5 (E) 10

243) Let n be a positive integer such that when we add 1 to the productof any n consecutive positive integers, the result is a perfect square, then nequals?

(A) 2 (B) 3 (C) 4 (D) 5 (E) none of these

244) Let f (x) = |x − p| + |x − 15| + |x − p − 15|, where 0 < p < 15.

Determine the minimum value taken by f (x) for x in the interval p ≤ x ≤ 15.

which comes out as (18+6−1) C6−1 =23 C5 . But this will also include the num-bers whose digit is greater than 9. which is not possible. So, we subtractthose solutions. Let a = 10 + a0 then a + b + c + d + e + f = 18 becomesa0 + b + c + d + e + f = 8. number of solutions is 13 C5 . Now in place of a, wecan have b, c, d, e and f so 6.13 C5

total steps A = 30 + 30 = 60!

214) (x2 + y 2 + z 2 + t2 )2 = x4 + y 4 + z 4 + t4

So after expanding, we see that in every pair of numbers, at least one mustbe 0. Therefore, 3 of the numbers must be 0. Then by the first equation, the4th number is 3. These yield the 4 values of s = 32008 , 32009 , 32010 , 32011

217) We need to use the fact that the medians of a triangle divide the triangleinto three triangles of equal area. this is used to find the area outside theoctagon. Please note that the octagon is not regular.Lets assume the area of the square is 9 Then the area of unshaded region is(2/3).8 − (1/8)4 = 29/6. subtracting we get area of S = 9 − 29/6 = 25/6 soratios is 25/54

218) Let x = 18! and y = 12! then lcm of xy 18

xy and 132 is what we need.The gcd of 18 and 132 is 6 so the lcm is 6 which equals (18!)(12!) xy 3! . henceabc = 18 × 12 × 3 = 648

219) clearly the remainder should be divisible by 9 but none of the optionsare so answer is e)

225) xy = (y + 1)4 clearly y = 4k

y = 1 will give a solutiony = 2 will give x = 9 another solution y = 4 will give x = 5 and y = 8 willgive x = 3 we get 4 solutions now we won’t get any solution because nextvalue is y = 12 but if y = 12 or greater x will become close to 1 but never 1so no more integral solution

226)3x2 − 8y 2 + 3x2 y 2 = 2008

15⇒ 3x2 (y 2 + 1) − 8y 2 = 2008 clearly rhs and the second term on lhs is divisibleby 8, so we need x as even and yas odd let x = 2k and y = 2m + 1 we get3k 2 (2m2 + 2m + 1) − (2m + 1)2 = 251clearly m=0 gives us a rhs divisible by 9 but lhs of 3clearly m=1 is not a solutionm=2 gives no solution chk divisibilitym = 3 gives k = 2m=4 no solutionm=5 nonem=6 gives noneclearly for a larger m we will get a smaller k which is not possibleso only one pair of (x, y) = (4, 7)

so x2 y = 112

227) One of my favourite questions√in this set. It √ uses some basic concepts √ 222 √rigorously. lets roll! See this T = ( x + y) + ( x − y)222 is clearly aninteger as the even powers will give integers and the odd powers which willmake surds will add up to zero√ due√to the presence of corresponding negativeterms. SO [T ] = T Let p = x − y now √ p > 1 or p < 1,if p < 1 then p222 is √very close to zero yet positive then [T ] = [( x+ y)222 ]+1 so √ option b is true 222 √ 222in this case if p > 1 then p will be very large So [T ] = [( x + y) ] + kwhere k is a positive integer much greater than 1 so option a and c are alwayswrong. So option e)

228) We have n = 100q + r, so n = 99q + q + r. Now, 11|(q + r), so we

have 11|n. so number of terms we need is just all 5 digit multiples of 11. thatis [99999/11] − [9999/11] = 8181

229) the min will be when p, q are as close as possible so p = 51 + x and

q = 51 − x now we have to make sure 51 + x, 51 − x both are primes whichcomes for x = 8 and 59 and 43 so p − q = 16

Simplifying, 73 ≥ 3x + 4y, so the largest possible value is 73 .

Testing possible values of x in terms of y, we see that x = y + 3 is the largestpossible value for positive x, y, so we have 6y + 9 = 7y − 5 ⇒ y = 14, so ouranswer is 294

234) Do we need a solution for this ? :) Anyways k 5 has same the unitdigit as k(check it). so last digit of n is 3 + 0 + 7 + 4mod10 = 4. All theoptions have 4 as unit digit.:) now clearly see lhs is divisible by 3. Now checkoptions n = 144, 174 are divisible by 3. But 174 is very large a number. hencen = 144

236) sin130o = sinOC

The angle x = 140o would put B right on top of A, so the solution thatwe want is x = 20o .

237) We can’t for sure say what is A but we can for sure say that A has3 elements 1, 2 and A

238) first we do it for just 3.

17there are 6 ∗ 5 ∗ 4 ∗ 3 ways for the first 3 to be different and the next one tobe the same. multiply by 4 c3 , then divide by 2 because we have one thatsthe same as another. thats 360 ∗ 2 = 720. divide by 64 = 1296. Then add6 ∗ 5 ∗ 4 ∗ 3 which is 360. so the total is 1080/1296 = 5/6

239) Since 4ABC = 3 · 4AOB = 6 · 4AON = 24 · 4OM Q. So the answer

241) From A to B there are clearly 3.2.1 = 6 paths. Now while comingback we cant take any of the edges again. So 2 paths are ruled out at Bitself. now out of the 4 remaining one will have to go through one edge usewhile coming to B. so we are left with 3 paths. So a total of 6.3 = 18 50242)xy = y 50 ⇒ x = y yclearly y =factors of 50 = 2.5.5 = 2.3 = 6 factors so 6 solutions here. Alsoy = 50k so y = 100 but no other value will give a integral x as k th root of50k will not turn out to be an integer for any other value of k. So a total of6 + 1 = 7 solutions

244) Using the two inequalites we are given about x and p, it becomesf (x) = (x − p) + (15 − x) + (15 + p − x) = 30 − x. This will be at a minimumwhen x is at its maximum which is when x = 15 so it is 30 − 15 = 15

3mod5. So we can divide 100 into four segments of 25 each, giving remain-der 3, 4, 2 and 1 on division by 5. Suppose we choose any athen we we canchoose b from the corresponding segment which makes the sum of the two

18 1remainders as 5. then probability is 100.25/(100.100) = 4Alternatively then the equation simplifies into 3a [1 + 3(b−a) ]. We can obvi-ously conclude that 3a is not a multiple of 5. Since a − b has equal probabilityof being anything, (assuming a >= b), we say that a − b = n. then sincethere is only one value (mod 5) where (3n )+1=0(mod 5), then we can deducethat n is such that 3n = 4mod5. Now clearly 0 ≤ n ≤ 99, again we can find 125 such n0 s out of 100 The probability of that happening is 4 ( 2n if n even246) Let g be the inverse of f . then g(n) = 2n − 1023 if n odd 5clearly k = g(g(g(g(g(k))))) = 2 k − 1023 ⇒ 31k = 1023 ⇒ k = 33 similarlym = 4m − 1023 ⇒ m = 341 so k + m = 341 + 33 = 374

19and 1. At each critical point there will be an increase of atleast one andat special points like1/4, 3/4 we will have increase of two and at 1/2, 1we have an increase of f our. So the numbers that can be represented are1, 2, 4, 5, 6, 10, 11, 12, 14, 15, 16, 20. So out of first 20 we get 12 so out of 1000we will get 600